Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)

Options

A. `9.0xx10^(-23)cm^(3)`
B. `6.023xx10^(-23)cm^(3)`
C. `3.xx10^(-23)cm^(3)`
D. `5.5xx10^(-23)cm^(3)`

Answer

C

Explanation

`"1 mole of "H_(2)O=6.02xx10^(23)" molecules"`
`=18 g = 18 cm^(3)`
`therefore" Volume occupied by 1 molecule of "H_(2)O`
`=(18)/(6.02xx10^(23))cm^(3)=3xx10^(-23)cm^(3)`